[proofplan]
We reduce the vector-valued representation to the scalar [Riesz Representation Theorem for Bounded Linear Functionals](/theorems/3037). For each coordinate $i = 1, \ldots, m$, we define a bounded linear functional $\Lambda_i$ on $C_0(\mathbb{R}^n)$ by embedding scalar functions into the $i$-th component, then apply the scalar Riesz theorem to obtain signed Radon measures $\mu_i$. The vector measure $\mu = (\mu_1, \ldots, \mu_m)$ represents $\Lambda$. For the norm identity, we establish both the upper bound $\|\Lambda\| \leq |\mu|(\mathbb{R}^n)$ using the definition of total variation and the lower bound by constructing test functions that approximate the polar decomposition $d\mu = \sigma \, d|\mu|$.
[/proofplan]
[step:Define the component functionals $\Lambda_i$ by embedding scalar functions into the $i$-th coordinate]
For each $i \in \{1, \ldots, m\}$, define the embedding map
\begin{align*}
\iota_i : C_0(\mathbb{R}^n) &\to C_0(\mathbb{R}^n; \mathbb{R}^m) \\
g &\mapsto (0, \ldots, 0, g, 0, \ldots, 0),
\end{align*}
where $g$ appears in the $i$-th component. Since $\|\iota_i(g)\|_\infty = \|g\|_\infty$, the map $\iota_i$ is a linear isometry. Define the linear functional
\begin{align*}
\Lambda_i : C_0(\mathbb{R}^n) &\to \mathbb{R} \\
g &\mapsto \Lambda(\iota_i(g)).
\end{align*}
Then $\Lambda_i$ is bounded with $\|\Lambda_i\| \leq \|\Lambda\| \cdot \|\iota_i\| = \|\Lambda\|$, since $\Lambda$ is bounded and $\iota_i$ is an isometry.
[/step]
[step:Apply the scalar Riesz theorem to each $\Lambda_i$ to obtain signed Radon measures $\mu_i$]
Each $\Lambda_i : C_0(\mathbb{R}^n) \to \mathbb{R}$ is a bounded linear functional on $C_0(\mathbb{R}^n)$. By the [Riesz Representation Theorem for Bounded Linear Functionals](/theorems/3037), there exists a unique signed Radon measure $\mu_i$ on $\mathbb{R}^n$ with $|\mu_i|(\mathbb{R}^n) < \infty$ such that
\begin{align*}
\Lambda_i(g) = \int_{\mathbb{R}^n} g \, d\mu_i
\end{align*}
for all $g \in C_0(\mathbb{R}^n)$, and $\|\Lambda_i\| = |\mu_i|(\mathbb{R}^n)$.
Define the $\mathbb{R}^m$-valued Radon measure $\mu = (\mu_1, \ldots, \mu_m)$. For any $f = (f_1, \ldots, f_m) \in C_0(\mathbb{R}^n; \mathbb{R}^m)$,
\begin{align*}
\Lambda(f) = \sum_{i=1}^m \Lambda_i(f_i) = \sum_{i=1}^m \int_{\mathbb{R}^n} f_i \, d\mu_i = \int_{\mathbb{R}^n} f \cdot d\mu.
\end{align*}
The total variation of $\mu$ satisfies $|\mu|(\mathbb{R}^n) \leq \sum_{i=1}^m |\mu_i|(\mathbb{R}^n) < \infty$, so $\mu$ has finite total variation.
[/step]
[step:Prove uniqueness of the representing measure]
Suppose $\nu = (\nu_1, \ldots, \nu_m)$ is another $\mathbb{R}^m$-valued Radon measure with finite total variation satisfying $\Lambda(f) = \int_{\mathbb{R}^n} f \cdot d\nu$ for all $f \in C_0(\mathbb{R}^n; \mathbb{R}^m)$. Taking $f = \iota_i(g)$ for arbitrary $g \in C_0(\mathbb{R}^n)$ gives
\begin{align*}
\int_{\mathbb{R}^n} g \, d\mu_i = \Lambda_i(g) = \int_{\mathbb{R}^n} g \, d\nu_i
\end{align*}
for all $g \in C_0(\mathbb{R}^n)$. By uniqueness in the scalar [Riesz Representation Theorem for Bounded Linear Functionals](/theorems/3037), $\mu_i = \nu_i$ for each $i$. Hence $\mu = \nu$.
[/step]
[step:Establish the upper bound $\|\Lambda\| \leq |\mu|(\mathbb{R}^n)$]
For any $f \in C_0(\mathbb{R}^n; \mathbb{R}^m)$ with $\|f\|_\infty \leq 1$, the Cauchy--Schwarz inequality in $\mathbb{R}^m$ gives $|f(x) \cdot \sigma(x)| \leq |f(x)| \cdot |\sigma(x)| \leq 1$ for any $\sigma(x)$ with $|\sigma(x)| \leq 1$. By the polar decomposition of the vector measure, $d\mu = \sigma \, d|\mu|$ where $\sigma : \mathbb{R}^n \to \mathbb{R}^m$ is a $|\mu|$-measurable function with $|\sigma(x)| = 1$ for $|\mu|$-a.e. $x$. Therefore
\begin{align*}
|\Lambda(f)| = \left|\int_{\mathbb{R}^n} f \cdot d\mu\right| = \left|\int_{\mathbb{R}^n} f \cdot \sigma \, d|\mu|\right| \leq \int_{\mathbb{R}^n} |f \cdot \sigma| \, d|\mu| \leq \|f\|_\infty \cdot |\mu|(\mathbb{R}^n).
\end{align*}
Taking the supremum over all $f$ with $\|f\|_\infty \leq 1$ gives $\|\Lambda\| \leq |\mu|(\mathbb{R}^n)$.
[/step]
[step:Establish the lower bound $\|\Lambda\| \geq |\mu|(\mathbb{R}^n)$ by approximating the polar direction]
Let $\varepsilon > 0$. Since $|\mu|$ is a finite Radon measure and $\sigma$ is $|\mu|$-measurable, [Lusin's Theorem](/theorems/3015) applies: there exists a compact set $K_\varepsilon \subset \mathbb{R}^n$ with $|\mu|(\mathbb{R}^n \setminus K_\varepsilon) < \varepsilon$ such that $\sigma|_{K_\varepsilon}$ is continuous. By the Tietze extension theorem, extend $\sigma|_{K_\varepsilon}$ to a continuous function $\tilde{\sigma} : \mathbb{R}^n \to \mathbb{R}^m$ with $\|\tilde{\sigma}\|_\infty \leq 1$. Multiplying by a continuous cutoff function $\psi \in C_c(\mathbb{R}^n)$ with $0 \leq \psi \leq 1$ and $\psi = 1$ on $K_\varepsilon$, define $g = \psi \tilde{\sigma} \in C_0(\mathbb{R}^n; \mathbb{R}^m)$. Then $\|g\|_\infty \leq 1$ and
\begin{align*}
\Lambda(g) = \int_{\mathbb{R}^n} g \cdot \sigma \, d|\mu| \geq \int_{K_\varepsilon} |\sigma|^2 \, d|\mu| - \int_{\mathbb{R}^n \setminus K_\varepsilon} |g \cdot \sigma| \, d|\mu|.
\end{align*}
On $K_\varepsilon$, $g = \tilde{\sigma} = \sigma$, so $g \cdot \sigma = |\sigma|^2 = 1$ $|\mu|$-a.e. For the second integral, $|g \cdot \sigma| \leq 1$. Therefore
\begin{align*}
\Lambda(g) \geq |\mu|(K_\varepsilon) - |\mu|(\mathbb{R}^n \setminus K_\varepsilon) \geq |\mu|(\mathbb{R}^n) - 2\varepsilon.
\end{align*}
Since $\|g\|_\infty \leq 1$, we have $\|\Lambda\| \geq \Lambda(g) \geq |\mu|(\mathbb{R}^n) - 2\varepsilon$. As $\varepsilon > 0$ was arbitrary, $\|\Lambda\| \geq |\mu|(\mathbb{R}^n)$.
[/step]
[step:Combine the bounds to conclude the norm identity]
From the upper bound $\|\Lambda\| \leq |\mu|(\mathbb{R}^n)$ and the lower bound $\|\Lambda\| \geq |\mu|(\mathbb{R}^n)$, we conclude
\begin{align*}
\|\Lambda\|_{\mathcal{L}(C_0(\mathbb{R}^n;\mathbb{R}^m), \mathbb{R})} = |\mu|(\mathbb{R}^n).
\end{align*}
This completes the proof: $\mu = (\mu_1, \ldots, \mu_m)$ is the unique $\mathbb{R}^m$-valued Radon measure with finite total variation that represents $\Lambda$, and the operator norm equals the total variation.
[/step]
